i have a list of directories in the current directory named with there permission codes (exemple : 552, 700, 777). I want to get the code permission from the name of directory and apply it to the directory and all the files it contains.

I tried with the xargs command : find . -name "[0-9][0-9][0-9]" -type d | xargs chmod -R [0-9][0-9][0-9]

the problem with this command it takes the first directory and its change the permission code of all directories.

├── 555
│   └── logs
│       ├── 01.log
│       ├── 02.log
│       ├── 03.log
│       ├── 04.log
│       ├── 05.log
│       ├── 06.log
│       └── 07.log
├── 700
│   └── data
│       └── data1.data

what I want : I have the 555 directory so I want to change all sub files and directory with permission code 555 and for the second directory I want to change all the subfiles and directory with the permission code 700

what my command do: it change all other files and subdirectories with the permission code of the first file 500

CodePudding user response：

Try

find . -name "[0-9][0-9][0-9]" -type d  | sed 's#\./\(.*\)#\1#' | xargs -I{} chmod -R {} {}

• the find is the same as yours.
• the sed is added to remove the ./ from the directory name. Find returns ./700, ./555, ...
• xargs with -I uses {} to reuse what it received into the command. So it says "chmod -R DIRNAME DIRNAME". So chmod -R 700 700 and so on.
• In your attempt, xargs chmod -R [0-9][0-9][0-9], there is nothing to link the [0-9] in the find to the [0-9] in xargs.

CodePudding user response：

Without xargs

find . -type d -regextype sed -regex ".*/[0-9]\{3\}$"| awk -F"/" '{print "chmod -R " $NF,$0}'|sh